Ignoring subpolynomial improvements, there are only two algorithms for this problem.
The first is a brute force over all triples and its running time is \(O(|A|\cdot |B| \cdot |C|)\) .
In the symmetric setting where \(|A|=|B|=|C|=n/3\)  this is \(O(n^3)\)  and it is optimal under both 3-SUM and APSP conjectures, as long as \(W=\Omega (n^3)\) 
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The second algorithm is faster when \(W\)  is small enough: It applies the standard exponentiation trick (encoding \(w\)  as \(2^w\) ) to reduce summation to multiplication and then uses fast matrix multiplication. In the symmetric setting the running time is \(O(W \cdot n^\omega )\)  and otherwise it is a complicated expression that depends on the rectangular matrix multiplication exponent.
Assuming \(\omega =2\) , the upper bound simplifies to \((W \cdot N)^{1+o(1)}\)  where \(N= (|A|\cdot |C| + |A|\cdot |B| + |B|\cdot |C|)\)  is an upper bound on the size of the graph.
It is natural to conjecture that these bounds cannot be broken for Zero-Triangle.
